# Still points in a rigid body

I'm reading Woodhouse's book on classical mechanics and I got stuck on this problem:

A rigid body has angular velocity $$\vec{\omega}$$ and has one point $$O$$ ﬁxed relative to a frame $$\tilde{R}$$. Show that if $$\vec{\omega} \times \tilde{D}\vec{\omega} \neq 0$$, then $$O$$ is the only point with zero acceleration relative to $$\tilde{R}$$.

Note. $$\tilde{D}\vec{\omega}$$ is the derivative with respect to time of $$\vec{\omega}$$ in $$\tilde{R}$$.

My approach. Suppose there exists $$P$$ in the rigid body with null acceleration and let $$\vec{r}$$ be the vector going from $$O$$ to $$P$$. Since the body is rigid, we have that the velocity $$\vec{v}_P$$ with respect to $$\tilde{R}$$ satisfies $$\vec{v}_P = \vec{v}_O + \vec{\omega} \times \vec{r} = \vec{\omega} \times \vec{r}$$ On differentiation with respect to time we get $$0 = \tilde{D}\vec{v}_P = \tilde{D}(\vec{\omega} \times \vec{r}) = (\tilde{D}\vec{\omega} \times \vec{r}) + (\vec{\omega} \times \tilde{D}\vec{r}) = (\tilde{D}\vec{\omega} \times \vec{r}) + (\vec{\omega} \times \vec{v}_P)$$ From this point on every manipulation I tried to prove that $$\vec{r}$$ must be the zero vector got me nowhere.

Does anyone know how to proceed?

Take the scalar product of your expression with with $$\omega$$ to get $$0 = \omega\cdot (\dot \omega\times r)+ \omega\cdot (\omega\times v_P)$$ The last term is zero and $$0=\omega\cdot (\dot \omega\times r)= r\cdot (\omega\times \dot \omega)$$ So $$r$$ has no component parallel to $$(\omega\times \dot \omega)$$ and $$r$$ must lie in the plane defined by $$\omega$$ and $$\dot \omega$$.
From $$0 = (\dot \omega\times r)+ (\omega\times v)$$ and $$(\omega\times v)=\omega \times(\omega\times r)= (\omega\cdot r)\omega- r |\omega|^2$$ we see firtly that that $$\dot \omega \times r$$ is perpendicular to $$r$$ so taking the scalar product we get
$$0=(\omega\cdot r)^2-|r|^2||\omega|^2$$ so $$r$$ is parallel to $$\omega$$ --- i.e $$r=a\omega$$ and this makes $$v=(\omega\times r)=0$$. Once we have $$v=0$$ then we have
$$0=(\dot \omega\times r)= a(\dot \omega\times \omega)$$ Thus $$a$$ is zero. Hence (because $$\omega\ne 0$$) $$r$$ is zero